# Linear algebra. A proof

• Jan 22nd 2013, 08:45 AM
ThinkingNumbers24
Linear algebra. A proof
Hi

I want to prove
(ab)AB = (aA)(bB), in which a, b E R and matrix multiplication AB are defined.

Proof.
Let A =(r_ij), B = (s_ij)

(aA + bB)
= ((a * r_ij) + (b * s_ij))
= (a * r_ij + b * s_ij)
= (a * r_ji + b * s_ji)
= (a * r_ji) + (b * s_ji)
= aA + bB

But I don't know are there errors in a proof. ?
• Jan 22nd 2013, 09:20 AM
topsquark
Re: Linear algebra. A proof
I think you are working at this too hard. We know that m(AB) = (mA)B = A(mB) since scalar multiplication commutes with the A and B operations. So you have
(ab)(AB) = (abA)B = (aAb)B = (aA)(bB)

-Dan
• Jan 22nd 2013, 09:45 AM
ThinkingNumbers24
Re: Linear algebra. A proof
Thank you. I see, and a proof is that short? Can you explain it more, so I can understand that proof, please.
• Jan 22nd 2013, 10:54 AM
topsquark
Re: Linear algebra. A proof
Quote:

Originally Posted by ThinkingNumbers24
Thank you. I see, and a proof is that short? Can you explain it more, so I can understand that proof, please.

No real proof is needed. By definition we know that, for a scalar n and an operator N that scalar multiplication is defined as
nN = Nn...Any operator commutes with a scalar. So that gives (ab)AB = (abA)B. And the rest is simply moving b to the other side of A.

It occurs to me that you might actually have to prove aA = Aa from another list of assumptions?

-Dan
• Jan 22nd 2013, 11:16 AM
ThinkingNumbers24
Re: Linear algebra. A proof
Thank you again. I just have that I want to prove
(ab)AB = (aA)(bB), in which a, b E R and matrix multiplication AB are defined.
I thought that above can be proven without by definitions.

Well, indeed, I think more, I believe that I will understand your answers.